Q. The work done, $W$ during an isothermal process in which 1 mole of the gas expands from an initial volume $V_{1}$ to a final volume $V_{2}$ is given by ($R$=gas constant, $T$=temperature)
A) $R\left( V_{2} - V_{1} \right)\log_{e}\left( \frac{T_{1}}{T_{2}} \right)$
B) $R\left( T_{2} - T_{1} \right)\log_{e}\left( \frac{V_{2}}{V_{1}} \right)$
C) $RT{\log_{e}(}{\frac{V_{2}}{V_{1}})\ \ }$
D) $2RT\log_{e}\left( \frac{V_{1}}{V_{2}} \right)$
Explanation
(c) Work done during isothermal process in expanding volume of gas from$V_{1}\ to\ V_{2}\ is\ given\ by$
$\ \ \ \ \ \ \ \ \ \ \ \ \ W = \int_{V_{1}}^{V_{2}}{p\ dV}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \int_{V_{1}}^{V_{2}}{\left( \frac{nRT}{V} \right)dV}\ \ \ \ \ \ \left( as\ p = \frac{nRT}{V} \right)$
$= nRT\int_{V_{1}}^{V_{2}}{\frac{dV}{V}\ \ \ \ \ \ \ \ \ \ \ \ \ (as\ T = constant)}$
= $nRT\log_{e}\left( \frac{V_{2}}{V_{1}} \right)$
For expansion of 1 mole of gas, ie, n=1
$W = RT\log_{e}\left( \frac{V_{2}}{V_{1}} \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.